\documentclass{beamer}

\usefonttheme[onlymath]{serif}

\mode<presentation>
{
    \usetheme{Warsaw}
    \setbeamercovered{transparent}
}

\title{Distributed Entanglement}
\subtitle{Paper Review}
\author{Hu Jingcheng}
\institute{Zhejiang University}

\pgfdeclareimage[height=0.5cm]{university-logo}{./pics/zjulogo}
\logo{\pgfuseimage{university-logo}}

\AtBeginSubsection[]
{
    \begin{frame}<beamer>{Outline}
        \tableofcontents[currentsection,currentsubsection]
    \end{frame}
}


\begin{document}

\begin{frame}
    \titlepage
\end{frame}

\begin{frame}{Outline}
    \tableofcontents
\end{frame}

\section{Introduction}
\begin{frame}
    \frametitle{Motivation}
    Why study quantum entanglement?
    \begin{itemize}
        \item Counter-intuitive and fascinating!
        \item Better understanding leads to better utilizing!
    \end{itemize}
\end{frame}
\begin{frame}
    \frametitle{Pitvotal Problems}
    How to quantify entanglement?
    \begin{itemize}
        \item Two-Qubit Entanglement
        \item More Qubits Entanglement
        \item More than two energy levels \dots
    \end{itemize}

    What are the properties of quantum entanglement?
    \begin{itemize}
        \item Is there any intrinsic limitation?
        \item How entanglement is related?
    \end{itemize}
\end{frame}


\section{Main Results}
\subsection{Quantifying Entanglement of Two Qubits: Concurrence}
\begin{frame} {Spin-Flip Operation for Single Qubit}
For a single qubit $|\Phi\rangle$

Define "spin-flip" operation
$$
|\tilde{\Phi}\rangle \triangleq \sigma_y  |\Phi^\star \rangle 
$$
where the conjugation is taken in standard basis $\{|0\rangle,|1\rangle\}$.
Visualization: Reversing the Bloch vector!
 
$\implies$ taking the state to its orthogonal state! 

Thus doing "spin-flip"!
\end{frame}

\begin{frame}{Concurrence for Pure State}
Two qubits A, B in pure state $|\Phi\rangle$

Applying "spin-flip" operation on both qubits:
$$
|\tilde{\Phi}\rangle \triangleq (\sigma_y \otimes \sigma_y) |\Phi^\star \rangle 
$$
define \textbf{concurrence} for AB\cite{concurrence_review}: 
$$
C(\Phi) \triangleq |\langle \Phi | \tilde{\Phi} \rangle |
$$
\end{frame}

\begin{frame}{Explicit Formula For Pure State Concurrence}
In standard basis $\{|00\rangle,|01\rangle, |10\rangle,|11\rangle\}$, 
we write $|\Phi\rangle = (a,b,c,d)$, then we have:
\begin{align}\label{eq:1}
|\tilde{\Phi}\rangle = (-d^\star, c^\star, b^\star, -a^\star) \implies C(\Phi) = 2|ad-bc|
\end{align} 
the relation between concurrence and how two qubits are entangled becomes apparent:
$$
\begin{cases}
    C = 0 &\iff \text{not entangled at all} \\
    0<C<1 &\iff \text{partially entangled} \\
    C = 1 &\iff \text{completely entangled} \\ 
\end{cases}
$$
\end{frame}

\begin{frame}{Explicit Formula For Pure State Concurrence}
introduce density operator $\rho_{AB}$ and reduced density operator $\rho_A = tr_B(\rho_{AB})$. 
We now prove a useful formula, and the main ideas of the proof are highly representative.
\begin{theorem}
\begin{align}\label{eq:2}
    {C(\Phi)}^2 = 4 \det{\rho_A}, \text{if AB is pure state}
\end{align}
\end{theorem}
\begin{proof}
Write $|\Phi\rangle$ in standard basis, and repeated indices are summed:
\begin{align*}
|\Phi\rangle &= a_{ij}|ij\rangle, i,j = 0 \text{ or } 1 \\
\implies&\rho_{AB} = |\Phi\rangle \langle\Phi| = a_{ij}a^\star_{lm} |ij\rangle\langle lm|
\end{align*}
\end{proof}
\end{frame}

\begin{frame}{Explicit Formula For Pure State Concurrence}
    \begin{proof}
    Doing partial trace:
    $$
    \implies\rho_A = tr_B \rho_{AB} = a_{ij}a^\star_{lm} tr_B |ij\rangle\langle lm| = \sum_j a_{ij}a^\star_{lj}|i\rangle\langle l| 
    $$
    write $\rho_A$ explicitly in matrix form:
    $$
    \rho_A = 
    \begin{bmatrix}
        aa^\star+bb^\star & ac^\star+bd^\star \\ 
        a^\star c + b^\star d & cc^\star + dd^\star
    \end{bmatrix}
    $$
    Then according to \autoref{eq:1}, expand the expression and then we can prove this formula.
    \end{proof}
\end{frame}

\begin{frame}{Concurrence for Mixed State}

Similarly define "spin-flip" operation:
$$
\tilde\rho = (\sigma_y \otimes \sigma_y) \rho^\star (\sigma_y \otimes \sigma_y)
$$
both $\rho$ and $\tilde{\rho}$ are positive operator $\implies\rho\tilde{\rho}$, though not Hermitian, 
has only real and non-negative eigenvalues.

We denote the \textit{square root} of the eigenvalues in decreasing order $\lambda_1,\lambda_2,\lambda_3,\lambda_4$.
\begin{theorem}{Calculating mixed state concurrence\cite{concurrence_mixed_state}}
\begin{align}\label{eq:3}
C(\rho) = \max{\{0,\lambda_1-\lambda_2-\lambda_3-\lambda_4\}}
\end{align}
\end{theorem}

\end{frame}

\subsection{A Definition of Three-Way Entanglement: Residual Entanglement}

\begin{frame}{Establishing Inequality with $C_{AB}$ and $C_{AC}$}

Given a \textbf{pure state} of three qubits A, B, and C,
how is $C_{AB}$ is related to $C_{AC}$ ? We first provide our conclusion:
\begin{theorem}
\begin{align}
C^2_{AB} + C^2_{AC} \le C^2_{A(BC)}
\end{align}
\end{theorem}

\end{frame}

\begin{frame}{Explaining $C_{A(BC)}$'s Well-definedness}
Note that ABC is in pure state, which means:
\begin{align*}
\rho_{ABC} &= |\Phi\rangle \langle\Phi| = a_{ijk}a^\star_{lmn}|ijk\rangle \langle lmn|\\
\implies \rho_{BC} &= tr_A\rho_{ABC} = \sum_i a_{ijk}a^\star_{imn}|jk\rangle \langle mn| \\
&= a_{0jk}a^\star_{0mn}|jk\rangle \langle mn| +a_{1jk}a^\star_{1mk}|jk\rangle \langle mn|  \\
&= |\Phi_1\rangle \langle\Phi_1| + |\Phi_2\rangle \langle\Phi_2|
\end{align*}
$\implies\dim \text{ran} (\rho_{BC}) = 2$, 
and $\text{ran} (\rho_{BC}) $ is spanned by $\{|\Phi_1\rangle,|\Phi_2\rangle\}$. 

$\implies$ at most two of the eigenvalues of the $\rho_{BC}$ is non-zero
$\implies$ We can equivalently treat BC as a single qubit in its mixed-state!
concurrence $C_{A(BC)}$ is well-defined!
\end{frame}

\begin{frame}{Sketch of the proof of $C^2_{AB} + C^2_{AC} \le C^2_{A(BC)}$}
\begin{proof}
$\rho_{AB}\tilde\rho_{AB}$ only has at most two non-zero eigenvalues denoting their square roots as $\lambda_1,\lambda_2$, 
and $\lambda_1\ge\lambda_2 \ge 0$. Then using \autoref{eq:3} we have:
\begin{align*}\label{eq:4}
C^2_{AB} &= {(\lambda_1 - \lambda_2)}^2  = \lambda_1^2+\lambda_2^2 - 2\lambda_1\lambda_2 \\
& = tr(\rho_{AB}\tilde\rho_{AB})- 2\lambda_1\lambda_2 \le tr(\rho_{AB}\tilde\rho_{AB})\\
\implies C^2_{AB}&+C^2_{BC} \le tr(\rho_{AB}\tilde\rho_{AB})+tr(\rho_{BC}\tilde\rho_{BC})
\end{align*}
Noticing that $\tilde\rho_{AB} = (\sigma_y \otimes \sigma_y) \rho^\star_{AB} (\sigma_y \otimes \sigma_y) = (\epsilon \otimes \epsilon) \rho^\star_{AB} (\epsilon \otimes \epsilon)$
where $\epsilon$ is 2-dimentional Levi-Civita tensor, and its matrix form is:
$$
\epsilon = 
\begin{bmatrix}
\epsilon_{00}&\epsilon_{01}\\\epsilon_{10}&\epsilon_{11} 
\end{bmatrix}
= \begin{bmatrix} 0&1\\-1&0 \end{bmatrix}
\implies\epsilon \otimes \epsilon = \epsilon_{ab}\epsilon_{cd}|ac\rangle \langle bd|
$$
\end{proof}
\end{frame}

\begin{frame}{Sketch of the proof of $C^2_{AB} + C^2_{AC} \le C^2_{A(BC)}$}
\begin{proof}
Expand the expression and after some algebra, we have the results:
\begin{align*}
tr(\rho_{AB}\tilde\rho_{AB}) &= 2(\det \rho_A+\det \rho_B-\det \rho_C)\\
\implies RHS = tr(\rho_{AB}\tilde\rho_{AB})&+tr(\rho_{AC}\tilde\rho_{AC}) = 4\det \rho_A
\end{align*}
Then using \autoref{eq:2}, we have:
$$
4\det \rho_A = C^2_{A(BC)}
$$
Thus finish the proof.
\end{proof}
\end{frame}


\begin{frame}{Residual Entanglement}{Using the Inequality to define Three-Way Entanglement}
According the inequality we just established, we can define \textit{resisual entanglement} $\tau_{ABC}$:
\begin{align*}
\tau_{ABC} \triangleq C^2_{A(BC)} - (C^2_{AB}+C^2_{BC}) = 2(\lambda^{AB}_1\lambda^{AB}_2 + \lambda^{AC}_1\lambda^{AC}_2) \ge 0
\end{align*}
We first give the results:
\begin{align*}
\tau_{ABC} &= C^2_{A(BC)} - (C^2_{AB}+C^2_{BC})= 4|d_1-2d_2+4d_3|\\
d_1 &= a^2_{000}a^2_{111}+a^2_{001}a^2_{110}+a^2_{010}a^2_{101}+a^2_{100}a^2_{011}\\
d_2 &= a_{000}a_{111}a_{011}a_{100} + a_{000}a_{111}a_{101}a_{010}\\
    &+ a_{000}a_{111}a_{110}a_{001}+ a_{011}a_{100}a_{101}a_{010}\\
    &+ a_{011}a_{100}a_{110}a_{001}+ a_{101}a_{010}a_{110}a_{001}\\
d_3 &= a_{000}a_{110}a_{101}a_{011}+a_{111}a_{001}a_{010}a_{100}
\end{align*}
\end{frame}

\begin{frame}{Invariance Under Permutation Transformation of $\tau_{ABC}$}
Visualize $a_{ijk}$ attached to the corner of a cube. 
Each term in $d_i$ is the product of four of the $a_{ijk}$ such that the "center of mass" of four is exactly the cube center, 
which can be exactly categorized into three class:
body diagonal and each used twice($d_1$), diagonal plane($d_2$), and two skew lines configuration($d_3$).
\begin{figure}[h]\label{fig:1}
\centering
\includegraphics[scale=0.3]{pics/cube.png}
\caption{Geometric Visualization of Terms in $d_i$}
\end{figure} 
\end{frame}

\begin{frame}{Sketch of Derivation of $\tau_{ABC}$}
\begin{proof}
\begin{align*}
\tau_{ABC} = 2(\lambda^{AB}_1\lambda^{AB}_2 + \lambda^{AC}_1\lambda^{AC}_2) \ge 0
\end{align*}
First evaluate the product $\lambda^{AB}_1\lambda^{AB}_2$.
If we consider the action of $\rho_{AB}\tilde\rho_{AB}$ restricted to its range, then $\lambda^{AB}_1\lambda^{AB}_2$ will be the 
square root of the determinant of this restricted transformation which we denote $R$, a $2\times2$ matrix.

Now we first give an explicit expression of $R_{ij}$.
Note that $span(\text{ran} (\rho_{AB})) = \{|\Phi_0\rangle,|\Phi_1\rangle\}$, 
where
$$
\begin{cases}
    |\Phi_0\rangle= a_{ij0}|ij\rangle \\ 
    |\Phi_1\rangle= a_{ij1}|ij\rangle \\ 
\end{cases}
$$
\end{proof}
\end{frame}

\begin{frame}{Sketch of Derivation of $\tau_{ABC}$}
\begin{proof}
Then we consider how $\rho_{AB}\tilde\rho_{AB}$ will transform $\{|\Phi_0\rangle,|\Phi_1\rangle\}$. 
Re-arange it into matrix form getting $R_{ij}$:
\begin{align}
    R_{ij} = a_{rqj}a^\star_{mni} \epsilon_{mm'}\epsilon_{nn'} a^\star_{m'n'p}a_{r'q'p}\epsilon_{r'r}\epsilon_{q'q}
\end{align}
Give the expression of $\det R$ directly
\begin{align}
\lambda^{AB}_1\lambda^{AB}_2 = \sqrt{\det R} = |d_1-2d_2+4d_3|
\end{align}
As the permutation-invariance discussion above, 

we can deduce $\lambda^{AB}_1\lambda^{AB}_2 =\lambda^{AC}_1\lambda^{AC}_2$:
\begin{align}
\implies \tau_{ABC} = 2(\lambda^{AB}_1\lambda^{AB}_2 + \lambda^{AC}_1\lambda^{AC}_2)=4|d_1-2d_2+4d_3|
\end{align}
\end{proof}
\end{frame}

\section{Further Discussions and Summary}

\begin{frame}{Generalization to Mixed State}
$C^2_{AB} + C^2_{AC} \le C^2_{A(BC)}$ still holds for situation where ABC is in mixed state. 
Now since ABC is in mixed state, we can no longer treat BC as a whole and define $C_{A(BC)}$ directly.
Let ABC in mixed state $\rho$, we have: 
$$
\langle C^2_{A(BC)} \rangle= \sum_i p_i C^2_{A(BC)}(\Phi_i)
$$
And define:
$$
C^2_{A(BC)}(\rho) = \inf \langle C^2_{A(BC)} \rangle
$$
Notice that the square of concurrence is what actually traded off in the inequality, 
therefore we propose our definition with the expectance of the square of concurrence. 
\end{frame}

\begin{frame}{Generalization to Mixed State}
Average the inequality on the corresponding decomposition of $\inf \langle C^2_{A(BC)} \rangle$, 
and utilize the convexity of concurrence on the set of density operators\cite{mixed_state_entanglement}:
\begin{align*}
    LHS &= C^2_{AB}(\rho) + C^2_{AC}(\rho)\\
        &= C^2_{AB}(\sum_i p_i\Phi_i) + C^2_{AC}(\sum_i p_i\Phi_i)\\
        &\le \sum_i p_i (C^2_{AB}(\Phi_i) + C^2_{AC}(\Phi_i) ) \\ 
        &\le \sum_i p_i C^2_{A(BC)}(\Phi_i)\\
        &= C^2_{A(BC)}(\rho) = RHS
\end{align*}
Therefore, we prove that the inequality still holds for mixed state of ABC.
\end{frame}

\begin{frame}{Summary and Outlook}
    Main results we obain:
    \begin{itemize}
        \item 2-qubit entanglement: concurrence
        \item An important inequality: $C^2_{AB} + C^2_{AC} \le C^2_{A(BC)}$
        \item Residual entanglement: $\tau_{ABC} = 4|d_1-2d_2+4d_3|$
    \end{itemize}
    Problems unsolved in the paper:
    \begin{itemize}
        \item More energy level?
        \item More qubits?
    \end{itemize}
    Some other interesting perspectives:
    \begin{itemize}
        \item Quantum Tensor Network\cite{tensor_network}
        \item \dots
    \end{itemize}
\end{frame}

\bibliographystyle{plain}
\bibliography{ref}
\end{document}